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Universal Gravitation - calculating orbital period of satellite

  • Thread starter EEristavi
  • Start date
  • #1
108
5

Homework Statement:

An artificial satellite circles the Earth in a circular orbit
at a location where the acceleration due to gravity is
9.00 m/s2. Determine the orbital period of the satellite.

Relevant Equations:

a = V^2/r
T = 2 Pi R / V

F = G M m / R^2

T^2 = 4 Pi^2 a^3 / (G M) - Kepler's third law
I didn't use Kepler's 3rd law and this may be the reason I have a wrong answer.
However, I want to know: where I make the mistake.

## ma = \frac {G M m} {R^2}##
## R = (\frac {G M} a)^{1/2}##

##a = \frac {v^2} R##
##V^2 = a R = a (\frac {G M} a)^{1/2} = (G a M)^{1/2}##
##V = (G a M)^{1/4}##

##T = \frac {2 \Pi R} V##
##T = \frac {2 \Pi (G M)^{1/2}} {(G M a)^{1/4} a^{1/2}} = \frac {2 \Pi(G M)^{1/4}} {a^{3/4}}##
 

Answers and Replies

  • #2
a = V^2/r
T = 2 Pi R / V

F = G M m / R^2

T^2 = 4 Pi^2 a^3 / (G M) - Kepler's third law
Actually the Kepler third law formula (if you meant "a" to be acceleration) is wrong. "a" should be R (radius).
 
  • #3
108
5
Actually the Kepler third law formula (if you meant "a" to be acceleration) is wrong. "a" should be R (radius).
I copied formula from textbook, it had meaning of radius - forgot to change to R it.
sorry
 
  • #4
Problem Statement: An artificial satellite circles the Earth in a circular orbit
at a location where the acceleration due to gravity is
9.00 m/s2. Determine the orbital period of the satellite.
Relevant Equations: a = V^2/r


##T = \frac {2 \Pi R} V##
##T = \frac {2 \Pi (G M)^{1/2}} {(G M a)^{1/4} a^{1/2}} = \frac {2 \Pi(G M)^{1/4}} {a^{3/4}}##
Well, I got the almost the same answer (almost 5400) for both methods. What you got?
 

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